2009 H2 Mathematics Paper 1 Question 2

Integration Techniques

Answers

p=2π3ln⁡3{p = \frac{2\pi}{3\ln 3}}

Full solutions

∫0114−x2  dx=[14ln⁡∣2+x2−x∣]01=14ln⁡3\begin{align*} & \int_0^1 \frac{1}{4 - x^2} \; \mathrm{d}x \\ & = \Bigg[ \frac{1}{4} \ln \left| \frac{2 + x}{2 - x} \right| \Bigg]_0^1 \\ & = \frac{1}{4} \ln 3 \end{align*}
∫012p11−p2x2  dx=1p∫012p11p2−x2  dx=1p[sin⁡−1(x1p)]012p=1p[sin⁡−1px]012p=1psin⁡−112−0=π6p\begin{align*} & \int_0^{\frac{1}{2p}} \frac{1}{\sqrt{1-p^2x^2}} \; \mathrm{d}x \\ & = \frac{1}{p} \int_0^{\frac{1}{2p}} \frac{1}{\sqrt{\frac{1}{p^2}-x^2}} \; \mathrm{d}x \\ & = \frac{1}{p} \left[ \sin^{-1} \left( \frac{x}{\frac{1}{p}} \right) \right]_0^{\frac{1}{2p}} \\ & = \frac{1}{p} \left[ \sin^{-1} px \right]_0^{\frac{1}{2p}} \\ & = \frac{1}{p} \sin^{-1} \frac{1}{2} - 0 \\ & = \frac{\pi}{6p} \end{align*}
π6p=14ln⁡3p=2π3ln⁡3  ■\begin{gather*} \frac{\pi}{6p} = \frac{1}{4} \ln 3 \\ p = \frac{2\pi}{3\ln 3} \; \blacksquare \end{gather*}